设函数f(x)=[cot(-x-π)sin(2π+x)]/[cos(-x)tan(3π-x)].(1)若f(α)=（根3

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设函数f(x)=[cot(-x-π)sin(2π+x)]/[cos(-x)tan(3π-x)].(1)若f(α)=（根3）/3,求α;(2)若cosα=α(|α|

【参考答案】
f(x)=[cot(-x-π)sin(2π+x)]/[cos(-x)tan(3π-x)]
=-cot(x+π)(sinx)/[(-cosx)tan(-x)]
=-cotxsinx/(cosxtanx)
=-(cosx/sinx)sinx/[cosx(sinx/cosx)]
=-cosx/sinx
=-cotx
(1)由f(a)=√3 /3得 cota=-√3 /3,故a=kπ- (π/3)
(2)由cosx=a得 sinx=±√(1-a²)
故f(x)=-cotx=-cosx/sinx=±a/√(1-a²)
有不理解的地方欢迎追问.

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